Results 1 to 3 of 3

Math Help - Irrational equation explanation

  1. #1
    Junior Member
    Joined
    Dec 2005
    Posts
    58

    Irrational equation explanation

    I have a problem again with irrational equation.

    I have to solve this equation:
     \sqrt[3]{x} + \sqrt[3]{{2x - 3}} = \sqrt[3]{{12(x - 1)}}

    I have solve it like this:
     \left( {\sqrt[3]{x} + \sqrt[3]{{2x - 3}}} \right)^3  = \left( {\sqrt[3]{{12(x - 1)}}} \right)^3
     x + 3\sqrt[3]{x}\sqrt[3]{{2x - 3}}(\sqrt[3]{x} + \sqrt[3]{{2x - 3}}) + 2x - 3 = 12x - 12
      \sqrt[3]{x}\sqrt[3]{{2x - 3}}(\sqrt[3]{x} + \sqrt[3]{{2x - 3}}) = 3x - 3
     \sqrt[3]{x}\sqrt[3]{{2x - 3}}\sqrt[3]{{12(x - 1)}} = 3x - 3
     \sqrt[3]{{x(2x - 3)(12x - 12)}} = 3x - 3
      x(2x - 3)4(3x - 3) = (3x - 3)^3
      4x(2x - 3) = (3x - 3)^2
     8x^2  - 12x = 9x^2  - 18x + 9
      x^2  - 6x + 9 = 0
      (x - 3)^2  = 0
    x = 3

    This is one solution but there is another one that is not obvious:  x=1.

    Now, I have come to that solution by assuming that right side of equation is 0: \sqrt[3]{{12(x - 1)}}=0

    Can someone explain me how to know exactly what are ALL solutions of irrational equations?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,804
    Thanks
    115
    Quote Originally Posted by DenMac21
    I have a problem again with irrational equation.

    I have to solve this equation:
     \sqrt[3]{x} + \sqrt[3]{{2x - 3}} = \sqrt[3]{{12(x - 1)}}

    I have solve it like this:
    [...snip...]

      x(2x - 3)4(3x - 3) = (3x - 3)^3
      4x(2x - 3) = (3x - 3)^2
    Hello,

    I've quoted those lines where you lost your 2nd solution.
      x(2x - 3)4(3x - 3) - (3x - 3)^3 =0
      (3x-3) \cdot \left( 4x(2x - 3) - (3x - 3)^2 \right) =0

    A product of two factors is zero, if one factor is zero:
    So you get:

      (3x-3)=0 \  \vee \  4x(2x - 3) - (3x - 3)^2  =0

    After solving both equations you've got all possible results.

    Bye
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2005
    Posts
    58
    Quote Originally Posted by earboth
    Hello,

    I've quoted those lines where you lost your 2nd solution.
      x(2x - 3)4(3x - 3) - (3x - 3)^3 =0
      (3x-3) \cdot \left( 4x(2x - 3) - (3x - 3)^2 \right) =0

    A product of two factors is zero, if one factor is zero:
    So you get:

      (3x-3)=0 \  \vee \  4x(2x - 3) - (3x - 3)^2  =0

    After solving both equations you've got all possible results.

    Bye
    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 19th 2010, 05:04 PM
  2. Replies: 2
    Last Post: January 31st 2010, 05:40 AM
  3. Laplacing ambigious equation, wants explanation:)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 17th 2009, 04:15 PM
  4. Replies: 7
    Last Post: January 29th 2009, 03:26 AM
  5. Irrational equation help
    Posted in the Algebra Forum
    Replies: 6
    Last Post: February 8th 2006, 06:31 PM

Search Tags


/mathhelpforum @mathhelpforum